Download - Enthalpy
MUET,SZAB Khairpur
K-13ME-05
AkashThahrani
FIRST-LAW ANALYSIS OF REACTING SYSTEMS
1st law of thermodynamics:
Reacting systems: the system in which chemical reaction take place
Steady-flow devices: operate for long periods of time under the same conditions.
Khairpur
Enthalpy of formation:
“ It can be viewed as the enthalpy of a substance at a specified state due to its chemical composition. “
Enthalpy of reaction or Enthalpy of Combustion:
It is the difference between theenthalpy of the products at a specifiedstate and the enthalpy of the reactantsat the same state for a completereaction.
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The enthalpy of a chemical component at a
specified state is the sum of
the enthalpy of the component at 25°C,
1atm(hf°), and the sensible enthalpy of the
Enthalpy of a chemical component
1atm(hf°), and the sensible enthalpy of the
component relative to 25 °C, 1 atm.
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When the changes in kinetic and potential energies are negligible,
The steady-flow energy balance relation for a chemically reacting steady-flow system can be written as:
FIRST-LAW ANALYSIS OF REACTING SYSTEMS
Steady-Flow Systems
Taking heat transfer to the system to be positive and work done by the system to be positive The energy balance relation just discussed can be expressed more compactly as:
_
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Most steady-flow combustion processes :
do not involve any work interactions.
involves heat output but no heat input
It expresses that :
“The heat output during a combustion process is simply the difference between theenergy of the reactants entering and the energy of the products leaving thecombustion chamber.”
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We note that
all the hydrogen in the fuel burns to H2O
but 10percent of the carbon burns incompletely and forms CO.
Also, the fuel is burned with excess air and thus there is some free O2 in the product gases.
The theoretical amount of air is determined from the Stoichiometric reaction to be:
Theoretical amount of air:
The theoretical amount of air is determined from the Stoichiometric reaction to be:
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Then the balanced equation for the actual combustion process with 50 percent excess air and some CO in the products becomes
(a) The air–fuel ratio for this combustion process is:
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(b) The heat transfer for this steady-flow combustion process is determinedfrom the steady-flow energy balance:
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Thus 363,880 kJ of heat is transferred from the combustion chamber for each kmol(44kg) of propane. This corresponds to 363,880/44 =8270 kJ of heat loss per kilogramof propane.
Then the rate of heat transfer for a mass flow rate of 0.05 kg/min for the propanebecomes
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Closed Systems:
The general closed-system energy balance relation:
for a stationary chemically reacting closed system as
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ENTHALPY OF FORMATION AND ENTHALPY OF COMBUSTION
Molecules of a system possess energy in various forms such as:
latent energy (associated with a change of state)
Chemical energy (associated with the molecular structure)
Nuclear energy (associated with the atomic structure)
Until now, we needed to dealwith were the sensible andlatent energies.
Now well consider chemicalenergy also.
The energy change of a system during a chemical reaction is due to:
Change in state Change in chemical composition.
Therefore, when the products formed during a chemicalreaction exit the reaction chamber at the inlet state of thereactants, we have
The energy change of the system in this case is due to thechanges in its chemical composition only.
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Enthalpy of reaction or Enthalpy of Combustion:
It is the difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction.
Consider the formation ofCO2 from its elements, carbonand oxygen, during a steady-flowcombustion process
The enthalpy of combustion isobviously a very useful propertyfor analyzing the combustionprocesses of fuels.
However, there are so manydifferent fuels and fuel mixturesthat it is not practical to list hcvalues for all possible cases.
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The enthalpy of combustion is obviously a very useful property for analyzing the combustion processes of fuels.
However, there are so many different fuels and fuel mixtures that it is not practical to list hC values for all possible cases.
Enthalpy of formation:
Besides, the enthalpy of combustion is not of much use when the combustion is incomplete.
Therefore a more practical approach would be to have a more fundamental property to represent the chemical energy of an element or a compound at some reference state.
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The enthalpy of formation istabulated for typical compounds.
The enthalpy of formation of theelements in their stable form istaken as zero.
The enthalpy of formation of theelements found naturally as
hfo
hfo
Substance Formula
M kJ/kmol
Air 28.97
0
Oxygen O2 32 0
Nitrogen N2 28 0
Carbon dioxide CO2 44 -393,520
Carbon monoxide CO 28 -110,530
Water (vapor) H2Ovap
18 -241,820elements found naturally asdiatomic elements, such asnitrogen, oxygen, and hydrogen, isdefined to be zero.
The enthalpies of formation forseveral combustion componentsare given in the following table.
p
Water (liquid) H2Oliq 18 -285,830
Methane CH4 16 -74,850
Acetylene C2H2 26 +226,730
Ethane C2H6 30 -84,680
Propane C3H8 44 -103,850
Butane C4H10 58 -126,150
Octane (vapor) C8H18 114 -208,450
Dodecane C12H26 170 -291,010
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Enthalpy of formation:
“ It can be viewed as the enthalpy of a substance at a specified state due to its chemical composition. “
Fundamental property to representFundamental property to representthe chemical energy of an element or acompound at some reference state.
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Solution: The enthalpy of combustion of a fuelis to be determined using enthalpy of formationdata:
for CO2 393,520 kJ/kmol
for H2O 285,830 kJ/kmolfor H2O 285,830 kJ/kmol
For C8H18 249,950 kJ/kmol
C8H18 + 12.5 O2 8CO2 + 9H2O
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